| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Lamina with hole removed |
| Difficulty | Standard +0.8 This is a two-part M3 centre of mass question requiring (a) integration to derive the standard cone result (3h/4 from vertex), which is a standard proof but requires careful setup of the integral with variable radius, and (b) applying the composite body formula with a removed cone, requiring correct handling of negative mass and careful arithmetic. The integration in part (a) and the composite calculation in part (b) both demand precision, making this moderately challenging but still within standard M3 scope. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Radius of element \(= \frac{x}{h}a\) | B1 | |
| Hence \(\int_0^h \pi\frac{x^2a^2}{h^2}x\,dx = \frac{1}{3}\pi a^2 h \times \bar{x}\) | M1 A1 | |
| \(\frac{1}{3}\pi a^2 h \times \bar{x} = \frac{\pi a^2}{h^2}\left[\frac{x^4}{4}\right]_0^h\) | M1 | |
| \(= \frac{\pi a^2 h^2}{4}\) | M1 | |
| \(\Rightarrow \bar{x} = \frac{3}{4}h\) | A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Volume of large cone \(= \frac{1}{3}\pi a^2 h = V\) | ||
| Volume of small cone \(= \frac{1}{3}\pi \times \frac{4a^2}{9} \times \frac{h}{2} = \frac{2}{9}V\) | ||
| Volume of \(S = \frac{7}{9}V\) | ||
| Volume: \(V\), \(\frac{2}{9}V\), \(\frac{7}{9}V\) | M1 A1 | |
| CM from \(A\): \(\frac{3}{4}h\), \(\frac{h}{2}+\frac{3}{4}\left(\frac{h}{2}\right)\), \(\bar{x}\) | B1 B1 | |
| \(V \times \frac{3}{4}h - \frac{2}{9}V\left(\frac{7h}{8}\right) = \frac{7}{9}V\bar{x}\) | M1 A1 | |
| \(\Rightarrow \bar{x} = \frac{5h}{7}\) | A1 | (7) |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Radius of element $= \frac{x}{h}a$ | B1 | |
| Hence $\int_0^h \pi\frac{x^2a^2}{h^2}x\,dx = \frac{1}{3}\pi a^2 h \times \bar{x}$ | M1 A1 | |
| $\frac{1}{3}\pi a^2 h \times \bar{x} = \frac{\pi a^2}{h^2}\left[\frac{x^4}{4}\right]_0^h$ | M1 | |
| $= \frac{\pi a^2 h^2}{4}$ | M1 | |
| $\Rightarrow \bar{x} = \frac{3}{4}h$ | A1 | (6) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Volume of large cone $= \frac{1}{3}\pi a^2 h = V$ | | |
| Volume of small cone $= \frac{1}{3}\pi \times \frac{4a^2}{9} \times \frac{h}{2} = \frac{2}{9}V$ | | |
| Volume of $S = \frac{7}{9}V$ | | |
| Volume: $V$, $\frac{2}{9}V$, $\frac{7}{9}V$ | M1 A1 | |
| CM from $A$: $\frac{3}{4}h$, $\frac{h}{2}+\frac{3}{4}\left(\frac{h}{2}\right)$, $\bar{x}$ | B1 B1 | |
| $V \times \frac{3}{4}h - \frac{2}{9}V\left(\frac{7h}{8}\right) = \frac{7}{9}V\bar{x}$ | M1 A1 | |
| $\Rightarrow \bar{x} = \frac{5h}{7}$ | A1 | (7) |
**(13 marks)**
---6. (a) Show, by integration, that the centre of mass of a uniform right cone, of radius $a$ and height $h$, is a distance $\frac { 3 } { 4 } h$ from the vertex of the cone.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{45c51316-7d58-4c16-9b5f-1d7421060a88-5_789_914_406_486}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
A uniform right cone $C$, of radius $a$ and height $h$, has vertex $A$. A solid $S$ is formed by removing from $C$ another cone, of radius $\frac { 2 } { 3 } a$ and height $\frac { 1 } { 2 } h$, with the same axis as $C$. The plane faces of the two cones coincide, as shown in Fig. 3.\\
(b) Find the distance of the centre of mass of $S$ from $A$.\\
\hfill \mbox{\textit{Edexcel M3 Q6 [14]}}